The dot product of $v=(v_1,v_2,v_3)$ and $w=(w_1,w_2,w_3)$ is simply
$$v\cdot w=v_1w_1+v_2w_2+v_3w_3.$$

In your problem, we have $w=(0,-2,-2)$. Computing $-2w$, we have
$$-2w=-2(0,-2,-2)=\bigl(-2\cdot0, -2\cdot(-2), -2\cdot(-2)\bigr)=(0,4,4).$$
Then
$$(-2w)\cdot w= (0,4,4)\cdot(0,-2,-2) =0\cdot0+(4)\cdot(-2)+(4)\cdot(-2)=-16.$$

Your expression at the end of your post demands that we find the norm of two vectors. The norm of $w$ is
$$\Vert w\Vert=\sqrt{ 0^2+(-2)^2+(-2)^2}=\sqrt {8}=2\sqrt2.$$
The norm of $-2w$ can be calculated in a similar manner; but you could also compute it using the rule $\Vert \alpha v\Vert=|\alpha|\Vert v\Vert$: